Question

If ${\log }_e\left(\frac{a+b}{2}\right)=\frac{1}{2}({\log }_{e}a+{\log }_{e}b),$ then relation between $a$ and $b$ will be.

Answer 1

${log}_e\left(\frac{a+b}{2}\right)=\frac{1}{2}({log}_{e}a+{log}_{e}b)$

$\Rightarrow {log}_e\left(\frac{a+b}{2}\right)=\frac{1}{2}{log}_{e}ab$ $(\because {log}_{e}m+{log}_{e}n={log}_{e}mn)$

$\Rightarrow {log}_e\left(\frac{a+b}{2}\right)={log}_e\sqrt{ab}$ $(\because \frac{1}{m}{log}_{e}n={log}_e{n}^{1/m})$

$\Rightarrow \frac{a+b}{2}=\sqrt{ab}$

Squaring,

$\Rightarrow {(a+b)}^2=4ab$

$\Rightarrow a^2+b^2+2ab=4ab$

$\Rightarrow a^2+b^2-2ab=0$

$\Rightarrow {(a-b)}^2=0$

$\Rightarrow a=b$